Double Integrals

A double integral generalizes the concept of a definite integral to functions of two variables, \( f(x, y) \). It's used to calculate the volume beneath the surface defined by \( z = f(x, y) \) and above a region \( D \) in the plane: \[\iint_D f(x, y)\, dA\] Here, \( dA \) denotes an infinitesimal element of area within region \( D \), and may be expressed as either \( dx\,dy \) or \( dy\,dx \), depending on the order of integration.

More broadly, the double integral \( \iint_D f(x, y)\, dA \) represents the volume of the solid bounded above by the surface \( z = f(x, y) \) and below by the region \( D \) lying in the \( xy \)-plane, where the lower surface is \( z = 0 \).

This geometric interpretation is valid as long as \( f(x, y) \geq 0 \) throughout the region \( D \).

However, if the function also takes on negative values, the double integral instead yields the net signed volume: the difference between the volume above the plane \( z = 0 \) (where \( f > 0 \)) and the volume below it (where \( f < 0 \)).

How to compute it

Double integrals are typically evaluated as iterated integrals, like so:

\[ \iint_D f(x, y)\, dA = \int_a^b \left( \int_{g_1(x)}^{g_2(x)} f(x, y)\, dy \right) dx \]

or with the order of integration reversed:

\[ \iint_D f(x, y)\, dA = \int_c^d \left( \int_{h_1(y)}^{h_2(y)} f(x, y)\, dx \right) dy \]

The choice of integration order depends on the shape of the region \( D \). Sometimes it's more convenient to integrate with respect to \( y \) first, and other times with respect to \( x \).

In short, you're free to choose the order that simplifies the computation, provided the region \( D \) is accurately described according to that order.

Note. Be aware that reversing the limits of integration (e.g., integrating in \( x \) first instead of \( y \)) may introduce a negative sign. It's essential to preserve the correct orientation of the limits to ensure the result has the proper sign.

A concrete example

Let’s look at the function \( f(x, y) = x + y \) over the rectangular region \( D = [0,1] \times [0,2] \).

The double integral becomes:

\[ \iint_D (x + y)\, dA = \int_0^1 \left( \int_0^2 (x + y)\, dy \right) dx \]

We start by integrating with respect to \( y \):

\[ \int_0^2 (x + y)\, dy = \left[ xy + \frac{y^2}{2} \right]_0^2 = 2x + 2 \]

Next, we integrate with respect to \( x \):

\[ \int_0^1 (2x + 2)\, dx = \left[ x^2 + 2x \right]_0^1 = 1 + 2 = 3 \]

So, the value of the double integral is 3. This represents the volume under the surface \( z = x + y \) and above the rectangle \( D = [0,1] \times [0,2] \) in the \( xy \)-plane.

Here's a 3D plot of the surface \( z = x + y \) over the rectangular region \( D = [0,1] \times [0,2] \):

The gray rectangle at the base represents the integration region \( D = [0,1] \times [0,2] \), and the colored solid between it and the surface corresponds to the volume of 3 computed via the double integral.

Example 2

Now let’s take the same function \( f(x, y) = x + y \), but integrate over a square region:

\[ D = [-1, 1] \times [-1, 1] \]

In this region, the function \( x + y \) behaves as follows:

• It's positive when \( x + y > 0 \),
• Negative when \( x + y < 0 \),
• And zero along the line \( x + y = 0 \).

Let’s compute the integral:

\[ \iint_D (x + y)\, dA = \int_{-1}^{1} \int_{-1}^{1} (x + y)\, dy\,dx \]

We begin by integrating with respect to \( y \):

\[ \int_{-1}^{1} (x + y)\, dy = \left[ xy + \frac{y^2}{2} \right]_{-1}^1 = x(1) + \frac{1}{2} - x(-1) - \frac{1}{2} = 2x \]

Now integrate with respect to \( x \):

\[ \int_{-1}^{1} 2x\, dx = \left[ x^2 \right]_{-1}^1 = 1 - 1 = 0 \]

Even though the function \( x + y \) isn't zero throughout the region, the double integral is zero because the positive and negative contributions perfectly cancel each other out.

This illustrates that a double integral doesn't always represent a physical volume; it can also express a net signed value - the result of subtracting the volume below the plane \( z = 0 \) from the volume above it.

It truly represents a volume only when the function remains non-negative (or non-positive) across the entire integration region.

And so on.